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How do you factor the following Homogenous Linear ODE with constant coefficients and what is the general solution:

$$L[f] = \left(\frac{\mathrm{d}}{\mathrm{d}x} +1\right)\left(\frac{\mathrm{d}}{\mathrm{d}x} +1\right)\left(\frac{\mathrm{d}^2 f}{\mathrm{d}x^2} + 4f\right) = 0$$

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  • $\begingroup$ What has me stumped is how to factor (d^2f/dx^2+4f)... $\endgroup$ – Greg Apr 17 '14 at 17:53
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You can factor the last bit using complex numbers: $$\left(\frac{\mathrm{d}^2 }{\mathrm{d}x^2} + 4\right)f = \left(\frac{\mathrm{d}}{\mathrm{d}x} +2i \right)\left(\frac{\mathrm{d}}{\mathrm{d}x} -2i\right)f $$ Hence, the general solution is a linear combination of terms $e^{-t}$, $te^{-t}$ (repeated root), $e^{2i t}$ and $e^{-2it}$. In many cases, it is desirable to have real-valued basis of solutions, so one uses $\cos 2t$ and $\sin 2t$ instead of $e^{2i t}$ and $e^{-2it}$.

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$$L=\left(\frac{\mathrm{d}}{\mathrm{d}x} +1\right)\left(\frac{\mathrm{d}}{\mathrm{d}x} +1\right)\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2} + 4\right)\\=\frac{{{d}^{4}}}{d {{x}^{4}}}+2 \left( \frac{{{d}^{3}}}{d {{x}^{3}}} \right) +5 \left( \frac{{{d}^{2}}}{d {{x}^{2}}} \right) +8 \left( \frac{d}{d x} \right) +4$$

General solution: $$f=C_1\sin(2x)+C_2\cos(2x)+C_3xe^{-x}+C_4e^{-x}$$

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